CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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$x \mapsto \sqrt{x}$ is a uniformly continuous bijection of $[0, \infty)$ onto itself whose inverse $x \mapsto x^{2}$ is not uniformly continuous

Statement refuted

Refuted claim: the inverse of a uniformly continuous bijection of metric spaces is uniformly continuous (Uniform continuity of a map of metric spaces: one $δ$ serving every point, Injection, surjection, bijection).

Let $X : = [0, \infty) \subseteq R$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length) with the metric inherited from the real line (The absolute value makes $R$ a metric space: $d (x, y) = ∣ x - y ∣$ is a metric, its open balls are the intervals $(x - r, x + r)$ , and it is unbounded, Isometry, isometric embedding, and the subspace metric on a subset), and let

$g : X \to X, g (x) : = \sqrt{x}, h : X \to X, h (x) : = x^{2} .$

Then $g$ is a bijection of $X$ onto itself with inverse $h$ ; $g$ is uniformly continuous; and $h$ is not uniformly continuous, the pairs $n + 1 / n$ and $n$ defeating every candidate $δ$ at $ε = 2$ .

Facts & Assumptions

Given: $X = [0, \infty)$ with the metric inherited from $R$ ; the maps $g (x) = \sqrt{x}$ and $h (x) = x^{2}$ ; a natural $n \geq 1$ ; reals $ε, δ > 0$ .

[L1]

Every $a \geq 0$ has a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ , Integer powers $a^{m}$ ).

[L2]

For $a, b \geq 0$ : $a \leq b$ if and only if $a^{2} \leq b^{2}$ (Squaring is monotone on the nonnegatives).

[L3]

$\sqrt{\cdot}$ on $[0, \infty)$ is $1 / 2$ -Hölder with constant $1$ and is uniformly continuous ( $\sqrt{\cdot}$ on $[0, \infty)$ is uniformly continuous and exactly $1 / 2$ -Hölder, and is not Lipschitz, Lipschitz map, $α$ -Hölder map for rational $0 < α \leq 1$ , and contraction, Contraction implies Lipschitz implies uniformly continuous implies continuous; every Hölder map is uniformly continuous, and a Lipschitz map on a bounded space is Hölder for every exponent).

[L4]

Factorisation of a difference of squares: $x^{2} - y^{2} = (x - y) (x + y)$ (Factorisation of $b^{n} - a^{n}$ , and the resulting Lipschitz estimate).

[L5]

For every real $η > 0$ there is a natural $n \geq 1$ with $1 / n < η$ ; positive naturals are positive reals; and $a > 0$ gives $1 / a > 0$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , Every complete ordered field is Archimedean, Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order).

[L6]

The absolute value and the usual metric of $R$ (Basic properties of the absolute value, The absolute value makes $R$ a metric space: $d (x, y) = ∣ x - y ∣$ is a metric, its open balls are the intervals $(x - r, x + r)$ , and it is unbounded, Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[L7]

Uniform continuity: one $δ > 0$ per $ε > 0$ serving every pair (Uniform continuity of a map of metric spaces: one $δ$ serving every point).

Counterexample

technique · direct

1.1

$g$ and $h$ map $X$ into $X$ : $\sqrt{x} \geq 0$ for $x \geq 0$ by [L1], and $x^{2} \geq 0$ for every real $x$ .

L1L2

1.2

$g$ is uniformly continuous on $X$ .

L3L7

1.3

For every real $t > 0$ and every $x, y \geq 0$ : $h (x) - h (y) = x^{2} - y^{2} = (x - y) (x + y)$ .

2.1

$h (g (x)) = (\sqrt{x})^{2} = x$ for every $x \geq 0$ , and $g (h (x)) = \sqrt{x^{2}} = x$ for every $x \geq 0$ , the latter because $x \geq 0$ and $x^{2} = x^{2}$ make $x$ the unique nonnegative square root of $x^{2}$ . So $g$ and $h$ are mutually inverse bijections of $X$ onto itself.

step 1.1L1L2

2.2

Fix a real $δ > 0$ and take a natural $n \geq 1$ with $1 / n < δ$ ; put $x : = n + 1 / n$ and $y : = n$ , both in $X$ . Then $∣ x - y ∣ = 1 / n < δ$ .

step 1.3L5L6

3.1

But $h (x) - h (y) = (x - y) (x + y) = \frac{1}{n} (2 n + \frac{1}{n}) = 2 + \frac{1}{n^{2}} > 2$ . So the pair $(x, y)$ satisfies $∣ x - y ∣ < δ$ and $∣ h (x) - h (y) ∣ > 2$ .

step 1.3step 2.2L5L6

4.1

Since $δ > 0$ was arbitrary, no $δ$ witnesses the uniform continuity condition for $h$ at $ε = 2$ , so $h$ is not uniformly continuous.

step 2.2step 3.1L7

5.1

Therefore $g$ is a uniformly continuous bijection of $X$ onto itself whose inverse $h$ is not uniformly continuous, which refutes the claim above.

step 2.1step 1.2step 4.1∎

Remarks

Both maps are continuous, and one direction of the pair is even Hölder. What fails is only the uniformity of the inverse: $h$ stretches distances by the factor $x + y$ , which is unbounded on $[0, \infty)$ , so no single $δ$ can serve every pair. On any bounded piece $[0, M]$ the same factor is at most $2 M$ , so $h$ is Lipschitz there and the phenomenon disappears (Lipschitz map, $α$ -Hölder map for rational $0 < α \leq 1$ , and contraction).
The witnesses shrink and their images do not. The pairs $n + 1 / n$ and $n$ are at distance $1 / n$ , which tends to $0$ , while their images stay more than $2$ apart. That is the shape of every failure of uniform continuity: a family of pairs whose separation vanishes and whose image separation does not.
Indexing. The witnesses are indexed by naturals $n \geq 1$ , because $1 / n$ appears; $N$ contains $0$ in this library and $1 / 0$ does not exist.
A homeomorphism can be uniformly continuous in one direction only. So "uniformly homeomorphic" is a genuinely stronger relation than "homeomorphic", and this pair shows the two differ; the metric-level version of the same point is the gap between topological and uniform equivalence (Topologically, uniformly and Lipschitz equivalent metrics on a set, On $(0, \infty)$ the metrics $∣ x - y ∣$ and $∣ 1 / x - 1 / y ∣$ share their topology and not their Cauchy sequences).

Depends on

Used by

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Dependency tree · two levels

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Sources

Uniform continuity (Wikipedia) (standard reference, not scraped)
Nth root (Wikipedia) (standard reference, not scraped)

x↦xx↦x​ is a uniformly continuous bijection of [0,∞)[0,∞) onto itself whose inverse x↦x2x↦x2 is not uniformly continuous